If I dissolve a Tums tablet #("1000 mg CaCO"_3)# in #"25 mL"# of water, how do I calculate the #"pH"# of the resultant suspension? When done experimentally, it was between #9# and #10#

There are several important things to keep in mind here. For starters, calcium carbonate is considered insoluble in water.

A saturated solution of calcium carbonate will hold about #"0.013 g"# of dissociated salt for every #"1 L"# of water at #25^@"C"# #-># see here.

So assuming that your solution is at #25^@"C"#, you can say that it will be able to hold

#25 color(red)(cancel(color(black)("mL water"))) * "0.013 g CaCO"_3/(10^3color(red)(cancel(color(black)("mL water")))) = "0.000325 g CaCO"_3#

of dissociated calcium carbonate. Use the molar mass of calcium carbonate to convert this to moles

#0.000325 color(red)(cancel(color(black)("g"))) * "1 mole CaCO"_3/(100.09 color(red)(cancel(color(black)("g")))) = 3.25 * 10^(-6) quad "moles CaCO"_3#

Now, the calcium carbonate that will dissociate will do so to produce calcium cations and carbonate anions in #1:1# mole ratios.

#"CaCO"_ (3(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + "CO"_ (3(aq))^(2-)#

This means that the saturated solution of calcium carbonate will contain #3.25 * 10^(-6)# moles of carbonate anions in #"25 mL"# of water, which is essentially equivalent to saying in #"25 mL"# of the solution.

The molarity of the carbonate anions will be equal to

#["CO"_3^(2-)] = (3.25 * 10^(-6)quad "moles")/(25 * 10^(-3) quad "L") = 1.30 * 10^(-4) quad "M"#

A second important thing to keep in mind here is that the carbonate anions will act as a weak base and react with water to form bicarbonate anions and hydroxide anions

#"CO"_ (3(aq))^(2-) + "H"_ 2"O"_ ((l)) rightleftharpoons "HCO"_ (3(aq))^(-) + "OH"_ ((aq))^(-)#

The acid dissociation constant for the bicarbonate anions is equal to--see here.

#K_ ("a HCO"_ 3^(-)) = 4.8 * 10^(-11)#

which means that the base dissociation constant of the carbonate anions, #K_ ("b CO"_ 3^(2-))#, will be

#K_ ("b CO"_ 3^(2-)) = (1 * 10^(-14))/(4.8 * 10^(-11)) = 2.08 * 10^(-4)#

This is the case because an aqueous solution at #25^@"C"# has #K_a * k_b = 1 * 10^(-14)#.

By definition, the base dissociation constant is equal to

#K_ ("b CO"_ 3^(2-)) = (["HCO"_3^(-)] * ["OH"^(-)])/(["CO"_3^(2-)])#

If you take #x# #"M"# to be the equilibrium concentration of hydroxide anions, you can say that at equilibrium, the solution will contain

#["HCO"_3^(-)] = ["OH"^(-)] = x quad "M"#

and

#["CO"_3^(2-)] = (1.30 * 10^(-4) - x) quad "M"#

This is the case because in order for the reaction to produce #x# #"M"# of hydroxide anions and of bicarbonate anions, it must consume #x# #"M"# of carbonate anions.

Plug this into the expression you have for the base dissociation constant to get

#K_ ("b CO"_ 3^(2-)) = (x * x)/(1.30 * 10^(-4) - x)#

#2.08 * 10^(-4) = x^2/(1.30 * 10^(-4) -x)#

Rearrange to quadratic equation form

#x^2 + 2.08 * 10^(-4) * x - 2.08 * 1.30 * 10^(-8) = 0#

This quadratic equation will produce two solutions, one positive and one negative, but since #x# is supposed to represent concentration, you can discard the negative one and say that

#x = 9.06 * 10^(-5)#

This means that, at equilibrium, the resulting solution has

#["OH"^(-)] = 9.06 * 10^(-5) quad "M"#

Finally, the #"pH"# of the solution will be equal to

#"pH" = 14 - [-log(["OH"^(-)])]#

#"pH"= 14 + log(9.06 * 10^(-5))#

#color(darkgreen)(ul(color(black)("pH" = 9.96)))#

I'll leave the answer rounded to two decimal places, the number of sig figs you have for the volume of water.